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If $\alpha ,\beta$ are roots of $x^{2}+2x+5= 0$ then the equation whose roots are $\left ( \alpha ^{4}-12\alpha \right )$ and $\left ( \beta ^{3}+\beta -8 \right )$ is

Option 1)
$x^{2}+7x+10= 0$

Option 2)
$x^{2}-7x+10= 0$

Option 3)
$x^{2}-7x-10= 0$

Option 4)
$x^{2}+7x-10= 0$

Answers (1)

best_answer

$\because \: \alpha ,\beta$ are roots, So $\alpha ^{2}=-2\alpha -5$ and $\beta ^{2}=-2\beta -5$ .

$\alpha ^{4}-12\alpha =\left ( a^{2} \right )^{2}-12\alpha =\left ( -2\alpha -5 \right )^{2}-12\alpha =4\alpha ^{2}+20\alpha +25-12\alpha =4\alpha ^{2}+8\alpha +25=4\left ( -2\alpha -5 \right )+8\alpha +25=5$

$\beta ^{3}+\beta -8=\beta \left ( -2\beta -5 \right )+\beta -8=-2\beta ^{2}-4\beta -8=-2\left ( -2\beta -5 \right )-4\beta -8=2$

$\therefore$ roots are 2 & 5

So equation is $x^{2}-7x+10=0$

$\therefore$ Option (B)

Degree reduction -

If $\alpha$ is a root of $ax^{2}+bx+c= 0$ then $a\alpha^{2}+b\alpha+c= 0\Rightarrow a\alpha ^{2}= -b\alpha -c$ which can be used to reduce degree of $\alpha$ from $2$ to $1$ .

-

Option 1)

$x^{2}+7x+10= 0$

This is incorrect

Option 2)

$x^{2}-7x+10= 0$

This is correct

Option 3)

$x^{2}-7x-10= 0$

This is incorrect

Option 4)

$x^{2}+7x-10= 0$

This is incorrect

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divya.saini

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